Is $f: \mathbb Q_{>0} \to \mathbb Q, f(x)=x^2$ an open map? Suppose $f: \mathbb{Q_{>0}} \to \mathbb{Q} $ maps every $x$ in its domain to $x^2$. And by an open map I mean a function which maps open sets to open sets. Given the natural topology on $\mathbb{Q}$, prove or disprove: $f$ is an open map.
Edit: I think $f$ cannot be open because in the range of $f$ we don't have the numbers whose square roots are irrational. But I don't know how to say this formally.
2nd edit: I think since a base of the topology on $\mathbb{Q}$ consist of open intervals, it is suffices to show that every open interval in $\mathbb{Q}$ has a point whose square root is not rational.
 A: A similar, but alternate view using sequences (we are blessed with sequences since we have usual topology on $\mathbb Q$, with usual metric):
Let $f:\mathbb Q _{>0} \to \mathbb Q$ be as above, $f(x) = x^2$. We show that it is not an open map by showing its image $f(\mathbb Q_{>0})$ is not an open set in $\mathbb Q$ (Note $\mathbb Q _{>0}$ is open in $\mathbb Q _{>0}$!). Denote the image $W = f(\mathbb Q_{>0}) = \{x^2 : x\in \mathbb Q_{>0}\}$.
Showing $W$ is not open in $\mathbb Q$ is equivalent to showing its complement $\mathbb Q - W$ is not closed in $\mathbb Q$. Since we have a natural metric, to show $\mathbb Q - W$ is not closed in $\mathbb Q$ is to exhibit a sequence whose terms are in $\mathbb Q - W$ and it converges in $\mathbb Q$ but the limit is not in $\mathbb Q - W$ (i.e. the limit is in $W$). [This is because closed sets contain all its limit points.]
Indeed, consider the sequence $(a_n)$ given by $a_n = n / (n + 1)$, $n\in \mathbb N$. Note that each $a_n$ is a rational number but is not a square of a rational number [you might want to prove this. I have edited to give one below]. Hence $a_n \in \mathbb Q - W$. But the limit of $(a_n)$ exists in $\mathbb Q$, namely $a_n \to 1$. Finally, note that $1\in W$. Thus $(a_n)$ is a convergent sequence in $\mathbb Q - W$ with limit in $W$, that is, $\mathbb Q - W$ is not closed. 
Hence $W$ is not open, and that $f$ is not an open map. ////
[Proof of $n/(n+1)$ not a square of rational number, for $n\in\mathbb N$:
 Suppose to the contrary that $n/(n+1) = a^2 / b^2$, for some $a,b,\in \mathbb N$ with $gcd(a,b) =1$, $b>a$. Then we have $b^2 n = a^2n +a^2$, which implies $n = a^2/(b^2 - a^2) = a^2/(b+a)(b-a)$, an integer. This shows that $(b+a)|a^2$, say $a^2 = x(b+a) = xb + xa$, for some integer $x\ge 1$. But this means $a^2 - xa = a(a-x) = xb$, or $a|xb$. But since $gcd(a,b)=1$, we must have $a|x$, say $x = ya$, some integer $y\ge 1$. Then by substitution, we see that $a^2 = ya(b+a) = yab +ya^2$, namely $a^2(1-y) = yab\le 0$, a contradiction.] (Does anyone else have an alternate proof to this "simple" statement?)
A: Let now $O \subseteq \mathbb Q_{>0}$ be open w.r.t. the subspace topology induced by $\mathbb R$. Then $O = U \cap \mathbb Q_{>0}$, where $U \subseteq \mathbb R$ is open in $\mathbb R$. Further, if we define $g: \mathbb R_{>0} \to \mathbb R, g(x) = x^2$ and $V := g(U)$, we obtain
$$
f(O) = g(O) = g(U \cap \mathbb Q_{>0}) = (g^{-1})^{-1}(U \cap \mathbb Q_{>0}) = g(U) \cap g(\mathbb Q_{>0}) = V \cap g(\mathbb Q_{>0}).
$$
Assume the latter set would be of the form $W \cap \mathbb Q$, where $W$ is open in $\mathbb R$. Then
$$
V \cap g(\mathbb Q_{>0}) = W \cap \mathbb Q.
$$
But if we choose $y \in \mathbb R$ and $\epsilon > 0$ such that $(y - \epsilon, y + \epsilon) \subseteq W$, we can find a rational $q$ in that interval which has irrational root, and thus $q \in W \cap \mathbb Q$, but not $q \in V \cap g(\mathbb Q_{>0})$. Thus, the answer is no; the map is not open.
EDIT: Let me also prove that $(z - \epsilon, z + \epsilon)$ contains a rational number which has irrational root (for arbitrary $z \in \mathbb R$). Clearly, it contains a rational $x/y$. If that rational has irrational root, we are done. Otherwise, we multiply it by a number of the form
$$
\frac{p}{p+1},
$$
where $p$ does not occur in the prime factorisation of the numerator and is sufficiently large such that the resulting number is still in the interval.
The resulting number's root is irrational by uniqueness of the prime decompositions of numerator and denominator of $(px)/((p+1)y)$ (if the root was rational, we would get a second decomposition by squaring the root).
